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Topic: 3 JEE matrix problems (Read 1213 times) |
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towr
wu::riddles Moderator Uberpuzzler
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On behalf of navdeep1771 I'm posting the follow problems from the 2010 Joint Entrance Exam
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« Last Edit: Jul 14th, 2018, 1:09pm by towr » |
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navdeep1771
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Re: 3 JEE matrix problems
« Reply #1 on: Jul 15th, 2018, 4:08am » |
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I was unable to post this image file. As it requires some minimum number of posts. So towr posted this from my side. Thanks towr Well these 3 problems are new to everyone (including towr).
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towr
wu::riddles Moderator Uberpuzzler
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Re: 3 JEE matrix problems
« Reply #2 on: Jul 15th, 2018, 7:18am » |
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I hadn't heard of skew-symmetric matrices before. But from the definition I'd say it's either a red herring or we're missing a statement that we're operating modulo p. On the other hand, it doesn't really matter. For odd p, a has to be 0 in a skew-symmetric matrix, because only then a=-a, and b = -c, so det(Tp)=b^2, which isn't divisible by p. Now if we try p=5, the only answer that fits turns out to be D: 2p-1
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navdeep1771
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Re: 3 JEE matrix problems
« Reply #3 on: Jul 18th, 2018, 12:30am » |
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That's Correct. Well mathematical proof goes like this:- We must have a^2 − b^2 = kp ⇒ (a + b) (a − b) = kp ⇒ either a − b = 0 or a + b is a multiple of p when a = b; number of matrices is p and when a + b = multiple of p ⇒ a, b has p − 1 ∴ Total number of matrices = p + p − 1 = 2p − 1. But what about '43' and '44'?
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towr
wu::riddles Moderator Uberpuzzler
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Re: 3 JEE matrix problems
« Reply #4 on: Jul 19th, 2018, 10:55pm » |
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Since a < p and p is odd, then 2a is only divisible by p when a = 0 If a= 0 then det (A) is a multiple p only if b=0 or c=0, which are 2p - 1 cases This seems irrelevant, because it's exactly what's not asked in either question 43 or 44 However, together with 43 and 44 it covers all possible A, so answer(43) + answer(44) + (2p - 1) must be p^3 Which means the answers to 43 and 44 should be C and D
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« Last Edit: Jul 19th, 2018, 10:57pm by towr » |
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navdeep1771
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Re: 3 JEE matrix problems
« Reply #5 on: Jul 20th, 2018, 8:15am » |
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@towr You are insane.
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